If $$cosec \theta = \frac{(x^2 + y^2)}{(x^2 - y^2)}$$, then what will be the value of $$\tan \theta$$ ?

Given, $$cosec \theta = \frac{(x^2 + y^2)}{(x^2 - y^2)}$$

$$\sin\theta=\frac{x^2-y^2}{x^2+y^2}$$

$$\cos\theta=\sqrt{1-\sin^2\theta\ }$$

$$=\sqrt{1-\left(\frac{x^2-y^2}{x^2+y^2}\right)^2}$$

$$=\sqrt{1-\frac{\left(x^2-y^2\right)^2}{\left(x^2+y^2\right)^2}}$$

$$=\sqrt{\frac{\left(x^2+y^2\right)^2-\left(x^2-y^2\right)^2}{\left(x^2+y^2\right)^2}}$$

$$=\sqrt{\frac{x^4+y^4+2x^2y^2-x^4-y^4+2x^2y^2}{\left(x^2+y^2\right)^2}}$$

$$=\sqrt{\frac{4x^2y^2}{\left(x^2+y^2\right)^2}}$$

$$=\frac{2xy}{x^2+y^2}$$

$$\therefore\ $$ $$\tan\theta\ =\frac{\sin\theta\ }{\cos\theta\ }=\frac{\frac{x^2-y^2}{x^2+y^2}}{\frac{2xy}{x^2+y^2}}=\frac{x^2-y^2}{2xy}$$

Hence, the correct answer is Option B